Difference between revisions of "2020 AMC 10A Problems/Problem 15"
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The probability that the divisor chosen is a perfect square is <cmath>\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m + n = 1 + 22 = 23 \implies \boxed{\textbf{(E) } 23 }</cmath> | The probability that the divisor chosen is a perfect square is <cmath>\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m + n = 1 + 22 = 23 \implies \boxed{\textbf{(E) } 23 }</cmath> | ||
− | ~mshell214 | + | ~mshell214, edited by Rzhpamath |
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==Video Solution== | ==Video Solution== |
Revision as of 22:20, 31 January 2020
Contents
Problem
A positive integer divisor of is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as , where and are relatively prime positive integers. What is ?
Solution
The prime factorization of is . This yields a total of divisors of In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that and can not be in the prime factorization of a perfect square because there is only one of each in Thus, there are perfect squares. (For , you can have , , , , , or 0 s, etc.) The probability that the divisor chosen is a perfect square is
~mshell214, edited by Rzhpamath
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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